Homework Help: Math

Posted by Halle on Saturday, June 13, 2009 at 2:38pm.

Hi there
I need help with the restrictions on the variables of this question:

Simplify. State any restrictions on the variables.

log(x^2+7x+12)/log(x^2-9)

So my answer is: log(x+4/x-3) which is correct.

Now for the restrictions, I have:
x<-4 and x>3

However the back of the book says the answer is only x>3 and not x<-4 for the restrictions of the variables. But when I sub in any number such as -5 into the ORIGINAL log equation, i get a defined and positive log of x value. So? is mine correct or theirs? Explain fully.

Math - Halle, Saturday, June 13, 2009 at 3:34pm
Sorry the question is:

log[(x^2+7x+12)/x^2-9)]
Math - Reiny, Saturday, June 13, 2009 at 3:50pm
first of all
log(x^2+7x+12)/log(x^2-9) is NOT equal to
log((x+4)/(x-3)) like you say
If the back of your book has that answer, then they are definitely WRONG

log [(x^2+7x+12)/(x^2-9)]
would be equal to log((x+4)/(x-3))

to find the restrictions you have to consider the numerator and denominator
remember we can only take logs of positive numbers
at the top:
log(x^2+7x+12) = log[(x+3)(x+4)]
x < -4 OR x > -3

at the bottom:
log(x^2 - 9) = log[(x-3)(x+3)]
x < -3 or x > 3

but both of these conditions must be met, so

x < -4 OR x > -3 AND x < -3 or x > 3

which leaves you with x < -4 or x > 3

notice that between -4 and +3 we get a negative for either one or the other function

another way to understand the restriction would be:
graph both y = (x+3)(x+4) and
y = (x+3)(x-3)
any x where either or both of those graphs dip below the x-axis would not be allowed.

so yes, they are wrong with their restrictions.
Math - Reiny, Saturday, June 13, 2009 at 3:51pm
ARGHHHH! saw your correction only after I posted.

So my answer would match your first posting, and it would clearly be a more interesting question.
Math - Reiny, Saturday, June 13, 2009 at 4:05pm
so now
log [(x+4)(x+3)/((x-3)(x+3))]
= log(x+4) + log(x+3) - log(x+3) - log(x-3)

now each of those terms must be defined.
so x>-4 AND x>-3 AND x>3

which is x > 3

notice that we canceled out log(x+3)
but for all x ≤ -3 that term would be undefined. So can we really do algebra with undefined numbers??

I know this is a bit tricky, since -5 (or any x < -4) would give a positive result in the original algebraic expression, but it would not work in the individuals terms.

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